Arithmetic - Average - Previous Year CAT/MBA Questions

Answer: 15

Text Explanation :

Answer: Option A

Text Explanation :

Answer: 70

Text Explanation :

Answer: Option C

Video Explanation

Text Explanation :

Let the average marks of a boy and girl be b and g respectively.

Given, (4g + 6b)/10 = 24
⇒ 4g + 6b = 240
⇒ 2g + 3b = 120   ...(1)

Also, b ≤ g ≤ 2b
⇒ 2b ≤ 2g ≤ 4b   
⇒ 5b ≤ 2g + 3b ≤ 7b  ...(2)

From (1) & (2), we get
​​​​​​​5b ≤ 120 ≤ 7b
⇒ b ≤ 24 and b ≤ 17(1/7)   ...(3)

Now we need to find integral values of 2g + 6b
= 2g + 3b + 3b
= 120 + 3b
120 + 3 × 120/7 ≤ 120 + 3b ≤ 120 + 3 × 24   ...from (3)
⇒ 171.42 ≤ 120 + 3b ≤ 192

∴ Integral possible values of 2g + 6b are from 172 till 192 i.e., 21 possible integral values.

Hence, option (c).

Answer: 16

Video Explanation

Text Explanation :

n people get average Rs. 352.
∴ Total amount distributed = 352n.

Now, 2 people get a total of 506 + 506 = Rs. 1012
⇒ Remaining amount = 352n - 1012

The average amount received by other is less than or equal to 330
∴ $\frac{352\mathrm{n}-1012}{\mathrm{n}-2}$ ≤ 330
⇒ 352n - 1012 ≤ 330n - 660
⇒ 22n ≤ 352
⇒ n ≤ 352/22 = 16
⇒ n ≤ 16

∴ n can take maximum value of 16.

Hence, 16.

Answer: Option C

Video Explanation

Text Explanation :

Let the old average be 'a'.
∴ Total weight of the three students = 3a

Let the weight of D and E be D kgs and E kgs respectively.

Because of D the average reduces by x kg.
⇒ a - x = (3a + D)/4
⇒ 4a - 4x = 3a + D
⇒ D = a - 4x   ...(1)

Because of E the average increases by 2x kg.
⇒ a + 2x = (3a + E)/4
⇒ 4a + 8x = 3a + E
⇒ E = a + 8x   ...(2)

We know, E - D = 12
⇒ (a + 8x) - (a - 4x) = 12
⇒ 12x = 12
⇒ x = 1

Hence, option (c).

Answer: Option B

Video Explanation

Text Explanation :

Let the number of students initially be I and the number of students joining be J.

Average of students initially be a kg, while those joining be (a + 3) kg such that final average of all students becomes (a + 0.6) kg

Initial Students        Students Joining
       I                                  J
       a                             a + 3
                    a + 0.6    
      2.4                             0.6

⇒ I/J = 2.4/0.6 = 4/1

Hence, option (b).

Answer: Option A

Video Explanation

Text Explanation :

Sum of the original 3 numbers = 3 × 13 = 39.

Now, $\frac{39+\mathrm{n}}{4}$ = odd = 2k - 1

⇒ 39 + n = 8k – 4

⇒ 43 + n = 8k

For k to be an integer least possible natural value of n = 5

Hence, option (a).
 

Answer: 14

Video Explanation

Text Explanation :

Given, $\frac{a_1+a_2+...+a_N}{N}$ = 300

⇒ a₁ + a₂ + ... + a_N = 300N            ...(1)

Also, $\frac{6a_1+a_2+...+a_N}{N}$ = 400

⇒ 6a₁ + a₂ + ... + a_N = 400N         ...(2)

(2) - (1)

⇒ 5a₁ = 100N

⇒ a₁ = 20N

Since, a₁ is the least of the given numbers it cannot be more than the average, hence a₁ ≤ 300.

⇒ N ≤ 15

If N = 1, a₁ = 20 and average cannot be equal to 300.

Hence, N can take all values from 2 till 15, i.e., 14 values.

Hence, 14.

Answer: 63

Video Explanation

Text Explanation :

Let the number of students in section B is x and that in A is (x – 10).
⇒ $\frac{(\mathrm{x}-10)\times 32+\mathrm{x}\times 60}{2\mathrm{x}-10}$ = integer

⇒ $\frac{92\mathrm{x}-320}{2\mathrm{x}-10}$ = integer

⇒ $\frac{46\mathrm{x}-160}{\mathrm{x}-5}$ = integer

⇒ $\frac{46(\mathrm{x}-5)+70}{\mathrm{x}-5}$ = integer

⇒ (x - 5) + $\frac{70}{\mathrm{x}-5}$ = integer

⇒ $\frac{70}{\mathrm{x}-5}$ should be an integer

∴ (x – 5) should be a factor of 70 while x > 10

⇒ Highest possible value of x = 75 while lowest possible value is 12

∴ Difference between highest and lowest values of x = 75 – 12 = 63.

Hence, 63.

Answer: Option B

Video Explanation

Text Explanation :

The average of lowest two number is 14, hence their sum = 28
The average of highest two number is 28, hence their sum = 56

To maximize the average of these six numbers, we need to maximize the middle two numbers. But it is also given that all the numbers are distinct.

Since all numbers are distinct out of the two highest numbers, one must be greater than 28 (a) and the other less than 28 (b).

To maximize the middle two numbers we need to maximize ‘b’. The maximum value ‘b’ can take is 27.

∴ Maximum value of the two middle numbers can be 25 and 26.

⇒ Maximum sum of the six numbers = 28 + 25 + 26 + 56 = 135
⇒ Maximum average of the six numbers = 135/6 = 22.5

Hence, option (b).

Answer: 35

Video Explanation

Text Explanation :

Let the number of patients in hospital A = x
∴ The number of patients in hospital B = x + 21

⇒ $\frac{200}{x}$ = $\frac{152}{x+21}$ + 3

⇒ 200x + 4200 = 152x + 3x² + 63x

⇒ 3x² + 15x - 4200 = 0

⇒ x² + 5x - 1400 = 0

⇒ x = 35 or -40 (not possible)

∴ Number of patients in hospital A = 35

Hence, 35.

Answer: Option C

Video Explanation

Text Explanation :

Let the amounts spent by the family each month be LCM (10, 20, 25, 50) = Rs. 100 for the first 3 months and then Rs. 50 for the next two months.

Amount of onion bought during month 1 = 100/10 = 10 kgs
Amount of onion bought during month 2 = 100/20 = 5 kgs
Amount of onion bought during month 3 = 100/25 = 4 kgs
Amount of onion bought during month 4 = 50/25 = 2 kgs
Amount of onion bought during month 5 = 50/50 = 1 kgs

∴ Total amount of onion bought = 10 + 5 + 4 + 2 + 1 = 22 kgs
Total amount spend on onions = 100 + 100 + 100 + 50 + 50 = Rs. 400.

∴ Average expense for onion per kg for these 5 months = 400/22 = 18.18 ≈ Rs. 18/kg.

Hence, option (c). 

Answer: 10

Video Explanation

Text Explanation :

Let the number of matches played so far be ‘b’ and the goals scored in these ‘n’ matches be ‘g’.

Now if 1 goal is scored in next 10 matches 

⇒ g + 1 = (n + 10) × 0.15

⇒ g = 0.15n + 1.5   …(1)

If 2 goals are scored in next 10 matches 

⇒ g + 2 = (n + 10) × 0.2

⇒ g = 0.2n + 1   …(2)

Solving (1) and (2) we get,

n = 10

Hence, 10.

Answer: 92

Video Explanation

Text Explanation :

Let the score of 5 toppers be x each.

To maximize the score of toppers we need to minimize the scores of the remaining 20 students.

∴ Score of remaining 25 students will be 30, 31, 32, … , 49.

⇒ 25 × 50 = 30 + 31 + … + 49 + 5x

⇒ 1250 = 790 + 5x

⇒ x = 92

Hence, 92.

Answer: Option D

Video Explanation

Text Explanation :

‘aabb’ is an even number hence, b can be either 0 or 2 or 4 or 6 or 8.

∴ ‘aabb’ can be:

1100, 1122, 1144, 1166, 1188

2200, 2222, 2244, 2266, 2288

…

9900, 9999, 9944, 9966, 9988

Adding all these number

= 5 × (1100 + ... + 9900) + 9 × (22 + 44 + 66 + 88) 

= (5 × 1100 × 45) + 9 × 22 × 10 = 5500 × 45 + 44 × 45 = 5544 × 45

∴ Average of all such numbers = (5544 × 45)/45 = 5544.

Hence, option (d).

Answer: Option C

Video Explanation

Text Explanation :

Given,

$a+\frac{b+c}{2}=5$

⇒ 2a + b + c = 10   …(1)

Also, $b+\frac{a+c}{2}=7$

⇒ 2b + a + c = 14   …(2)

Solving (1) and (2) we get,

b – a = 4

⇒ b = a + 4

Substituting this in the (1)

⇒ 2a + a + 4 + c = 10

⇒ 3a + c = 6

Given all three as positive integers, only possible value for a is 1. (c cannot be 0)

So, when a = 1, c = 3 and b = 5

∴ a + b = 6.

Hence, option (c).

Answer: Option A

Video Explanation

Text Explanation :

Let the highest and lowest score be h and l respectively and total score of remaining 8 students be x.

The mean of the lowest 9 scores is 42

⇒ x + l = 9 × 42 = 378   …(1)

The mean of the highest 9 scores is 47

⇒ x + h = 9 × 47 = 423   …(2)

(2) – (1) 

⇒ h – l = 423 – 378 = 45

Case 1: Least possible average is when we minimize the highest marks. The least highest marks can be 47.

∴ Lowest score = 47 – 45 = 2

∴ Least average = $\frac{9\times 47+2}{10}$ = 42.5

Case 2: Highest possible average is when we maximize the lowest marks. The highest lowest marks can be 42.

∴ Highest score = 42 + 45 = 87

∴ Highest average = $\frac{9\times 42+87}{10}$ = 46.5

∴ The required difference = 46.5 – 42.5 = 4

Hence, option (a).

Answer: Option D

Video Explanation

Text Explanation :

Average score in n innings is 30 and in n + 2 innings is 29. Score in last 2 innings is 38 and 15

∴ 30n + 38 + 15 = 29(n + 2)

⇒ n = 5

∴ Total runs score in first n innings = 5 × 30 = 150

Also, he scored less than 38 runs in each of these 5 innings.

To calculate the lowest possible score in an innings we try to maximize the score of other 4 innings. Maximum runs score in each of the 4 innings can be 37.

∴ x + 4 × 37 = 150

⇒ x = 2

Hence, option (d).

Answer: 18

Video Explanation

Text Explanation :

Let Tom's age to be 'x' years.

∴ Dick's age = 3x.

∴ Harry's age = 6x.

Given, Dick's age is 1 year less than the average age of all three,

∴ 3x = (x + 3x + 6x)/3 – 1

⇒ 9x = 10x – 3

⇒ x = 3

∴ Harry’s age = 6x = 6 × 3 = 18.

Hence, 18.

Answer: Option B

Video Explanation

Text Explanation :

Let marks of Gautam be G.

∴ G + (62 × 21) = T ...(I) (where T is the total marks of all 22 students)

82.5 + (21 × x) = T ...(II) (where x is the average marks of 21 students other than Ramesh) 

The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh.

∴ (T/22) = 1 + x ...(III)

Solving (I), (II) and (III), we get; x = 60.5, T = 1353 and G = 51.

Hence, option (b).

Answer: Option C

Video Explanation

Text Explanation :

Sum of all 30 integers = 30 × 5 = 150.

Exactly 20 of the 30 integers do not exceed 5 that means the remaining 10 integers are greater than 5.

To keep the average of the first 20 integers as high as possible, we need to keep the average of the remaining 10 integers ,which are above 5, as low as possible. Since we are dealing with integers, the least value that the 10 integers above 5 can take is 6.

So, the sum of the 10 integers = 10 × 6 = 60.

Hence, the sum of the remaining 20 integers = 150 - 60 = 90

∴ The average of the remaining 20 = 90/20 = 4.5

Hence, option (c).

Answer: 60

Video Explanation

Text Explanation :

Let there be ‘n’ tests and their average be ‘x’.

By the given conditions,

When first 10 tests are not considered:
⇒ $\frac{nx-200}{n-10}$ = x + 1

⇒ x = $\frac{nx-200}{n-10}$ - 1   ...(1)

When last 10 tests are not considered:
⇒ $\frac{nx-300}{n-10}$ = x - 1

⇒ x = $\frac{nx-300}{n-10}$ + 1   ...(2)

Equating (1) and (2)
∴ $\frac{nx-200}{n-10}$ - 1 = $\frac{nx-300}{n-10}$ + 1

∴ $\frac{nx-200}{n-10}$ - $\frac{nx-300}{n-10}$ = 2

∴ $\frac{100}{n-10}$ = 2

∴ n = 60

Hence, 60.

Answer: Option B

Video Explanation

Text Explanation :

Let x is average age of 30 people whose age is 51 years and above. Clearly, x ≥ 51

Let y is average age of 39 people whose age is less than 51 years.

∴ 30x + 39y = 69 × 38
⇒ 10x + 13y = 874

Now to maximise the value of y, we need to minimise the value of x. The least possible value of y = 51.

∴ 10 × 51 + 13y = 874
⇒ 13y = 874 - 510 = 364
⇒ y = 364/13 = 28

Thus, the largest possible average age of the people whose ages are below 51 years is 28 years.

Hence, option (b).

Answer: Option B

Video Explanation

Text Explanation :

Given numbers are a₁, a₂, a₃,..., a₅₁,100.

If the arithmetic mean of these 52 numbers is ‘x’, we have
a₁ + a₂ + a₃ + ... + a₅₁ + 100 = 52x   ...(1)

Given: the arithmetic mean of a₂, a₃, a₄,...,a₅₁, 100 is ‘x + 1’
a₂ + a₃ + ⋯ + a₅₁ + 100 = 51(x + 1) = 51x + 51   ...(2)

Solving (1) and (2), we get:
∴ a₁ = x − 51   ...(3)

Now for a₁ to be maximum possible, x has to be maximum possible.

For x to be maximum possible each of a₂, a₃, ..., till a₅₁ has to be maximum possible. But they are also distinct integers.
∴ Maximum value of
a₅₁ = 99 [it has to be an integer less than a₅₂]
a₅₀ = 98 [it has to be an integer less than a₅₁]
​​​​​​​...
a₂ = 50

⇒ a₂, a₃, ..., a₅₂ are consecutinve integers from 50 till 100.

⇒ Average of (a₂, a₃, ..., a₅₂) = 1/2 × (50 + 100) = 75 = x + 1 [From (2)]
⇒ x = 74

From (3)
⇒ a₁ = 74 - 51 = 23

Hence, option (b).